Calculus Examples

$\arctan (x + y) = y^{2} + \frac{π}{4}$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d x} (\arctan (x + y)) = \frac{d}{d x} (y^{2} + \frac{π}{4})$

Step 2

Differentiate the left side of the equation.

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Step 2.1

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \arctan (x)$ and $g (x) = x + y$ .

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Step 2.1.1

To apply the Chain Rule, set $u$ as $x + y$ .

$\frac{d}{d u} [\arctan (u)] \frac{d}{d x} [x + y]$

Step 2.1.2

The derivative of $\arctan (u)$ with respect to $u$ is $\frac{1}{1 + u^{2}}$ .

$\frac{1}{1 + u^{2}} \frac{d}{d x} [x + y]$

Step 2.1.3

Replace all occurrences of $u$ with $x + y$ .

$\frac{1}{1 + {(x + y)}^{2}} \frac{d}{d x} [x + y]$

Step 2.2

Differentiate.

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Step 2.2.1

By the Sum Rule, the derivative of $x + y$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [y]$ .

$\frac{1}{1 + {(x + y)}^{2}} (\frac{d}{d x} [x] + \frac{d}{d x} [y])$

Step 2.2.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$\frac{1}{1 + {(x + y)}^{2}} (1 + \frac{d}{d x} [y])$

Step 2.3

Rewrite $\frac{d}{d x} [y]$ as $y'$ .

$\frac{1}{1 + {(x + y)}^{2}} (1 + y')$

Step 2.4

Reorder the factors of $\frac{1}{1 + {(x + y)}^{2}} (1 + y')$ .

$(1 + y') \frac{1}{1 + {(x + y)}^{2}}$

Step 3

Differentiate the right side of the equation.

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Step 3.1

By the Sum Rule, the derivative of $y^{2} + \frac{π}{4}$ with respect to $x$ is $\frac{d}{d x} [y^{2}] + \frac{d}{d x} [\frac{π}{4}]$ .

$\frac{d}{d x} [y^{2}] + \frac{d}{d x} [\frac{π}{4}]$

Step 3.2

Evaluate $\frac{d}{d x} [y^{2}]$ .

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Step 3.2.1

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{2}$ and $g (x) = y$ .

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Step 3.2.1.1

To apply the Chain Rule, set $u$ as $y$ .

$\frac{d}{d u} [u^{2}] \frac{d}{d x} [y] + \frac{d}{d x} [\frac{π}{4}]$

Step 3.2.1.2

Differentiate using the Power Rule which states that $\frac{d}{d u} [u^{n}]$ is $n u^{n - 1}$ where $n = 2$ .

$2 u \frac{d}{d x} [y] + \frac{d}{d x} [\frac{π}{4}]$

Step 3.2.1.3

Replace all occurrences of $u$ with $y$ .

$2 y \frac{d}{d x} [y] + \frac{d}{d x} [\frac{π}{4}]$

Step 3.2.2

Rewrite $\frac{d}{d x} [y]$ as $y'$ .

$2 y y' + \frac{d}{d x} [\frac{π}{4}]$

Step 3.3

Differentiate using the Constant Rule.

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Step 3.3.1

Since $\frac{π}{4}$ is constant with respect to $x$ , the derivative of $\frac{π}{4}$ with respect to $x$ is $0$ .

$2 y y' + 0$

Step 3.3.2

Add $2 y y'$ and $0$ .

$2 y y'$

Step 4

Reform the equation by setting the left side equal to the right side.

$(1 + y') (\frac{1}{1 + {(x + y)}^{2}}) = 2 y y'$

Step 5

Solve for $y'$ .

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Step 5.1

Simplify $(1 + y') \frac{1}{1 + {(x + y)}^{2}}$ .

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Step 5.1.1

Rewrite.

$0 + 0 + (1 + y') \frac{1}{1 + {(x + y)}^{2}} = 2 y y'$

Step 5.1.2

Simplify by adding zeros.

$(1 + y') \frac{1}{1 + {(x + y)}^{2}} = 2 y y'$

Step 5.1.3

Multiply $1 + y'$ by $\frac{1}{1 + {(x + y)}^{2}}$ .

$\frac{1 + y'}{1 + {(x + y)}^{2}} = 2 y y'$

Step 5.2

Move all terms containing $y'$ to the left side of the equation.

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Step 5.2.1

Subtract $2 y y'$ from both sides of the equation.

$\frac{1 + y'}{1 + {(x + y)}^{2}} - 2 y y' = 0$

Step 5.2.2

Simplify the denominator.

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Step 5.2.2.1

Rewrite ${(x + y)}^{2}$ as $(x + y) (x + y)$ .

$\frac{1 + y'}{1 + (x + y) (x + y)} - 2 y y' = 0$

Step 5.2.2.2

Expand $(x + y) (x + y)$ using the FOIL Method.

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Step 5.2.2.2.1

Apply the distributive property.

$\frac{1 + y'}{1 + x (x + y) + y (x + y)} - 2 y y' = 0$

Step 5.2.2.2.2

Apply the distributive property.

$\frac{1 + y'}{1 + x \cdot x + x y + y (x + y)} - 2 y y' = 0$

Step 5.2.2.2.3

Apply the distributive property.

$\frac{1 + y'}{1 + x \cdot x + x y + y x + y \cdot y} - 2 y y' = 0$

Step 5.2.2.3

Simplify and combine like terms.

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Step 5.2.2.3.1

Simplify each term.

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Step 5.2.2.3.1.1

Multiply $x$ by $x$ .

$\frac{1 + y'}{1 + x^{2} + x y + y x + y \cdot y} - 2 y y' = 0$

Step 5.2.2.3.1.2

Multiply $y$ by $y$ .

$\frac{1 + y'}{1 + x^{2} + x y + y x + y^{2}} - 2 y y' = 0$

Step 5.2.2.3.2

Add $x y$ and $y x$ .

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Step 5.2.2.3.2.1

Reorder $y$ and $x$ .

$\frac{1 + y'}{1 + x^{2} + x y + x y + y^{2}} - 2 y y' = 0$

Step 5.2.2.3.2.2

Add $x y$ and $x y$ .

$\frac{1 + y'}{1 + x^{2} + 2 x y + y^{2}} - 2 y y' = 0$

Step 5.2.3

To write $- 2 y y'$ as a fraction with a common denominator, multiply by $\frac{1 + x^{2} + 2 x y + y^{2}}{1 + x^{2} + 2 x y + y^{2}}$ .

$\frac{1 + y'}{1 + x^{2} + 2 x y + y^{2}} - 2 y y' \cdot \frac{1 + x^{2} + 2 x y + y^{2}}{1 + x^{2} + 2 x y + y^{2}} = 0$

Step 5.2.4

Combine $- 2 y y'$ and $\frac{1 + x^{2} + 2 x y + y^{2}}{1 + x^{2} + 2 x y + y^{2}}$ .

$\frac{1 + y'}{1 + x^{2} + 2 x y + y^{2}} + \frac{- 2 y y' (1 + x^{2} + 2 x y + y^{2})}{1 + x^{2} + 2 x y + y^{2}} = 0$

Step 5.2.5

Combine the numerators over the common denominator.

$\frac{1 + y' - 2 y y' (1 + x^{2} + 2 x y + y^{2})}{1 + x^{2} + 2 x y + y^{2}} = 0$

Step 5.2.6

Simplify the numerator.

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Step 5.2.6.1

Apply the distributive property.

$\frac{1 + y' - 2 y y' \cdot 1 - 2 y y' x^{2} - 2 y y' (2 x y) - 2 y y' y^{2}}{1 + x^{2} + 2 x y + y^{2}} = 0$

Step 5.2.6.2

Simplify.

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Step 5.2.6.2.1

Multiply $- 2$ by $1$ .

$\frac{1 + y' - 2 y y' - 2 y y' x^{2} - 2 y y' (2 x y) - 2 y y' y^{2}}{1 + x^{2} + 2 x y + y^{2}} = 0$

Step 5.2.6.2.2

Multiply $y$ by $y$ by adding the exponents.

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Step 5.2.6.2.2.1

Move $y$ .

$\frac{1 + y' - 2 y y' - 2 y y' x^{2} - 2 (y \cdot y) y' (2 x) - 2 y y' y^{2}}{1 + x^{2} + 2 x y + y^{2}} = 0$

Step 5.2.6.2.2.2

Multiply $y$ by $y$ .

$\frac{1 + y' - 2 y y' - 2 y y' x^{2} - 2 y^{2} y' (2 x) - 2 y y' y^{2}}{1 + x^{2} + 2 x y + y^{2}} = 0$

Step 5.2.6.2.3

Multiply $y$ by $y^{2}$ by adding the exponents.

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Step 5.2.6.2.3.1

Move $y^{2}$ .

$\frac{1 + y' - 2 y y' - 2 y y' x^{2} - 2 y^{2} y' (2 x) - 2 (y^{2} y) y'}{1 + x^{2} + 2 x y + y^{2}} = 0$

Step 5.2.6.2.3.2

Multiply $y^{2}$ by $y$ .

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Step 5.2.6.2.3.2.1

Raise $y$ to the power of $1$ .

$\frac{1 + y' - 2 y y' - 2 y y' x^{2} - 2 y^{2} y' (2 x) - 2 (y^{2} y^{1}) y'}{1 + x^{2} + 2 x y + y^{2}} = 0$

Step 5.2.6.2.3.2.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\frac{1 + y' - 2 y y' - 2 y y' x^{2} - 2 y^{2} y' (2 x) - 2 y^{2 + 1} y'}{1 + x^{2} + 2 x y + y^{2}} = 0$

Step 5.2.6.2.3.3

Add $2$ and $1$ .

$\frac{1 + y' - 2 y y' - 2 y y' x^{2} - 2 y^{2} y' (2 x) - 2 y^{3} y'}{1 + x^{2} + 2 x y + y^{2}} = 0$

Step 5.2.6.3

Multiply $2$ by $- 2$ .

$\frac{1 + y' - 2 y y' - 2 y y' x^{2} - 4 y^{2} y' x - 2 y^{3} y'}{1 + x^{2} + 2 x y + y^{2}} = 0$

Step 5.3

Set the numerator equal to zero.

$1 + y' - 2 y y' - 2 y y' x^{2} - 4 y^{2} y' x - 2 y^{3} y' = 0$

Step 5.4

Solve the equation for $y'$ .

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Step 5.4.1

Subtract $1$ from both sides of the equation.

$y' - 2 y y' - 2 y y' x^{2} - 4 y^{2} y' x - 2 y^{3} y' = - 1$

Step 5.4.2

Factor $y'$ out of $y' - 2 y y' - 2 y y' x^{2} - 4 y^{2} y' x - 2 y^{3} y'$ .

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Step 5.4.2.1

Factor $y'$ out of ${y'}^{1}$ .

$y' \cdot 1 - 2 y y' - 2 y y' x^{2} - 4 y^{2} y' x - 2 y^{3} y' = - 1$

Step 5.4.2.2

Factor $y'$ out of $- 2 y y'$ .

$y' \cdot 1 + y' (- 2 y) - 2 y y' x^{2} - 4 y^{2} y' x - 2 y^{3} y' = - 1$

Step 5.4.2.3

Factor $y'$ out of $- 2 y y' x^{2}$ .

$y' \cdot 1 + y' (- 2 y) + y' (- 2 y x^{2}) - 4 y^{2} y' x - 2 y^{3} y' = - 1$

Step 5.4.2.4

Factor $y'$ out of $- 4 y^{2} y' x$ .

$y' \cdot 1 + y' (- 2 y) + y' (- 2 y x^{2}) + y' (- 4 y^{2} x) - 2 y^{3} y' = - 1$

Step 5.4.2.5

Factor $y'$ out of $- 2 y^{3} y'$ .

$y' \cdot 1 + y' (- 2 y) + y' (- 2 y x^{2}) + y' (- 4 y^{2} x) + y' (- 2 y^{3}) = - 1$

Step 5.4.2.6

Factor $y'$ out of $y' \cdot 1 + y' (- 2 y)$ .

$y' \cdot (1 - 2 y) + y' (- 2 y x^{2}) + y' (- 4 y^{2} x) + y' (- 2 y^{3}) = - 1$

Step 5.4.2.7

Factor $y'$ out of $y' \cdot (1 - 2 y) + y' (- 2 y x^{2})$ .

$y' \cdot (1 - 2 y - 2 y x^{2}) + y' (- 4 y^{2} x) + y' (- 2 y^{3}) = - 1$

Step 5.4.2.8

Factor $y'$ out of $y' \cdot (1 - 2 y - 2 y x^{2}) + y' (- 4 y^{2} x)$ .

$y' \cdot (1 - 2 y - 2 y x^{2} - 4 y^{2} x) + y' (- 2 y^{3}) = - 1$

Step 5.4.2.9

Factor $y'$ out of $y' \cdot (1 - 2 y - 2 y x^{2} - 4 y^{2} x) + y' (- 2 y^{3})$ .

$y' (1 - 2 y - 2 y x^{2} - 4 y^{2} x - 2 y^{3}) = - 1$

Step 5.4.3

Divide each term in $y' (1 - 2 y - 2 y x^{2} - 4 y^{2} x - 2 y^{3}) = - 1$ by $1 - 2 y - 2 y x^{2} - 4 y^{2} x - 2 y^{3}$ and simplify.

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Step 5.4.3.1

Divide each term in $y' (1 - 2 y - 2 y x^{2} - 4 y^{2} x - 2 y^{3}) = - 1$ by $1 - 2 y - 2 y x^{2} - 4 y^{2} x - 2 y^{3}$ .

$\frac{y' (1 - 2 y - 2 y x^{2} - 4 y^{2} x - 2 y^{3})}{1 - 2 y - 2 y x^{2} - 4 y^{2} x - 2 y^{3}} = \frac{- 1}{1 - 2 y - 2 y x^{2} - 4 y^{2} x - 2 y^{3}}$

Step 5.4.3.2

Simplify the left side.

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Step 5.4.3.2.1

Cancel the common factor of $1 - 2 y - 2 y x^{2} - 4 y^{2} x - 2 y^{3}$ .

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Step 5.4.3.2.1.1

Cancel the common factor.

$\frac{y' (1 - 2 y - 2 y x^{2} - 4 y^{2} x - 2 y^{3})}{1 - 2 y - 2 y x^{2} - 4 y^{2} x - 2 y^{3}} = \frac{- 1}{1 - 2 y - 2 y x^{2} - 4 y^{2} x - 2 y^{3}}$

Step 5.4.3.2.1.2

Divide $y'$ by $1$ .

$y' = \frac{- 1}{1 - 2 y - 2 y x^{2} - 4 y^{2} x - 2 y^{3}}$

Step 5.4.3.3

Simplify the right side.

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Step 5.4.3.3.1

Move the negative in front of the fraction.

$y' = - \frac{1}{1 - 2 y - 2 y x^{2} - 4 y^{2} x - 2 y^{3}}$

Step 6

Replace $y'$ with $\frac{d y}{d x}$ .

$\frac{d y}{d x} = - \frac{1}{1 - 2 y - 2 y x^{2} - 4 y^{2} x - 2 y^{3}}$